Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I prove that $ 10200300040000100004000300201$ is not a perfect square ? This number is divisible with $3$ only one time. Is it a good reason and it is enough ?

thanks :)

share|improve this question
11  
yes. If $3|n$ and $n$ is a perfect square, then $9|n$. –  N. S. Dec 28 '12 at 15:16

4 Answers 4

up vote 33 down vote accepted

Yes. Assume by contradiction that your number is a perfect square $n^2$.

Since $3|n \cdot n$ and $3$ is prime, it follows that $3|n$. Then $n=3k$ and hence $n^2=9k^2$. Thus, your number is divisible by $9$, contradiction.

share|improve this answer

Assume the distinct prime factors of $n$ are $p_1, p_2, \cdots, p_k$, as:

$$n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$

For $n$ to be a perfect square, a necessary and sufficient condition is $\alpha_1, \alpha_2, \cdots, \alpha_k$ even, such that $\sqrt{p_i^{\alpha_i}}$ be an integer $\implies$ $\sqrt{n}$ be an integer. Otherwise, $\sqrt{n}$ is irrational and hence $n$ is not a perfect square.


If $3$ divides $n$ only one time, then we must have:

$$n = 3^1 p_2^{\alpha_2} \cdots p_k^{\alpha_k}$$

$1$ is not even, therefore $n$ is not a perfect square.

share|improve this answer

HINT: $ 3|n \wedge (n=k^2,k\in\mathbb{N})$ follows that $9|n$.

HINT 2: Use the divisibility test of $9$.

share|improve this answer

The digital root of a square is $1, 4, 7$, or $9$. Yours is $3$...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.